Hard
Number of Submatrices That Sum to Target — C++
Full explanation · Time O(m^2 * n) · Space O(n)
// Time: O(m^2*n), m is min(r, c), n is max(r, c)
// Space: O(n), which doesn't include transposed space
class Solution {
public:
int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
if (matrix.size() > matrix[0].size()) {
const auto& transposed = transpose(matrix);
return numSubmatrixSumTarget(const_cast<vector<vector<int>>&>(transposed),
target);
}
for (int i = 0; i < matrix.size(); ++i) {
for (int j = 0; j < matrix[i].size() - 1; ++j) {
matrix[i][j + 1] += matrix[i][j];
}
}
int result = 0;
for (int i = 0; i < matrix.size(); ++i) {
vector<int> prefix_sum(matrix[i].size());
for (int j = i; j < matrix.size(); ++j) {
unordered_map<int, int> lookup;
++lookup[0];
for (int k = 0; k < matrix[i].size(); ++k) {
prefix_sum[k] += matrix[j][k];
if (lookup.count(prefix_sum[k] - target)) {
result += lookup[prefix_sum[k] - target];
}
++lookup[prefix_sum[k]];
}
}
}
return result;
}
private:
vector<vector<int>> transpose(const vector<vector<int>>& matrix) {
vector<vector<int>> result(matrix[0].size(),
vector<int>(matrix.size()));
for (int i = 0; i < matrix.size(); ++i) {
for (int j = 0; j < matrix[i].size(); ++j) {
result[j][i] = matrix[i][j];
}
}
return result;
}
};